Efficiently Evaluate a Multivariable Function on a List of Tuples

I have a function G[x,y,z,r,s,t] in the code below. Quite simply, I want to generate a list of all possible tuples $(x,y,z,r,s,t)$ such that all six entries are taken from the set $0,1,2,...,N-1$ where $N$ will be an integer parameter of the problem. I then want to evaluate $G$ on each of these tuples and as efficiently as I can, count the number of the tuples which evaluate to 0 modulo $N$. I at least have the following:

G[x,y,z,r,s,t]

G[x_, y_, z_, r_, s_, t_] :=
x*y*z*(r^3 + s^3 + t^3) - r*s*t*(x^3 + y^3 + z^3);
F[N_] := Range[0, N - 1];
Tup[N_] := Tuples[F[N], 6];

G[x_, y_, z_, r_, s_, t_] :=
x*y*z*(r^3 + s^3 + t^3) - r*s*t*(x^3 + y^3 + z^3);
F[N_] := Range[0, N - 1];
Tup[N_] := Tuples[F[N], 6];

So Tup[N] is a list of all my tuples of interest. I was about to do a "for loop" ranging over the number of tuples and evaluate something like

Tup[N]

G[Tup[2][[4]][[1]], Tup[2][[4]][[2]], Tup[2][[4]][[3]],
Tup[2][[4]][[4]], Tup[2][[4]][[5]], Tup[2][[4]][[6]]]

G[Tup[2][[4]][[1]], Tup[2][[4]][[2]], Tup[2][[4]][[3]],
Tup[2][[4]][[4]], Tup[2][[4]][[5]], Tup[2][[4]][[6]]]

for example, but this seems to be extremely inefficient. I'm sure there must be a smarter way! So given my G[x,y,z,r,s,t] as well as Tup[N] how can I construct a function P[N] which will output the number of tuples which evaluate to 0 modulo $N$?

G[x,y,z,r,s,t]

Tup[N]

P[N]

G @@@ Tup[10];

G @@ Transpose[Tup[10]];

2 Answers
2

Here are several ways to perform the computations along with timings:

G2[X_] := X[[1]] X[[2]] X[[3]] (X[[4]]^3 + X[[5]]^3 + X[[6]]^3) -
X[[4]] X[[5]] X[[6]] (X[[1]]^3 + X[[2]]^3 + X[[3]]^3);
cG2 = With[code = G2[Array[Compile`GetElement[X, #] &, 6]],
Compile[X, _Integer, 1,
code,
CompilationTarget -> "C",
RuntimeAttributes -> Listable,
Parallelization -> True,
RuntimeOptions -> "Speed"
]
];


data = Tup[10];
a = G @@@ data; // RepeatedTiming // First
b = G @@ Transpose[data]; // RepeatedTiming // First
c = G2 /@ data; // RepeatedTiming // First
d = cG2[data]; // RepeatedTiming // First
e = Flatten[Outer[G, ## & @@ ConstantArray[F[10], 6]]]; // RepeatedTiming // First
a == b == c == d == e


4.012

0.068

7.30

0.0312

3.80

True

a

b

b

d

a

c

c

a

tup1[N_] := Tuples[G @@ F[N], 6]


or

tup2[N_] := Flatten[Outer[G, ## & @@ ConstantArray[F[N], 6]]]


They both give the same result as G @@@ Tup[N] suggested by Henrik:

G @@@ Tup[N]

tup1[7] == tup2[7] == G @@@ Tup[7]


True

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